SSC CGL level Solution Set 75 Fractions decimal and indices 7

This is the 75th solution set of 10 practice problem exercise for SSC CGL exam and 7th on topic fractions, decimals and indices. A number of special time-saving techniques and methods are applied for quickly solving the problems in mind as far as possible.

Solution 1: Problem analysis and solving

When all four options are continued fractions, evaluating each to the fullest extent will take time. So we adopt the strategy of evaluating the choices with smaller depth, that is, first and third options in the first stage because these two can be evaluated faster than the othe two.

Technique of evaluation of continued fractions: Example - Option 3

As an example of the process of mentally evaluating a continued fraction let us take up the third choice,

By the nature of continued fractions, till the end result is reached, the result of each evaluation is inverted. This is the key pattern here.

Second step: invert result $\displaystyle\frac{9}{8}$ to $\displaystyle\frac{8}{9}$ and the expression to be evaluated now is, $2+\displaystyle\frac{8}{9}$.

The important point at this step is the value of 9 in the denominator of the second term. By the addition now, this value won't be changed and at the next step it will go up to the numerator. As our target numerator is 13, we can reject this choice.

We don't need to evaluate to the end.

Similarly the first choice ends up with a numerator value of 17 and is rejected.

Likewise, we get numerator value 17 again for the second choice and reject the choice.

Though we know by now that the fourth choice is the answer, still we check it quickly.

Problem 2.

When $\left(\displaystyle\frac{1}{2} -\displaystyle\frac{1}{4}+\displaystyle\frac{1}{5}-\displaystyle\frac{1}{6}\right)$ is divided by $\left(\displaystyle\frac{2}{5} -\displaystyle\frac{5}{9}+\displaystyle\frac{3}{5}-\displaystyle\frac{7}{18}\right)$ the result is,

$5\displaystyle\frac{1}{10}$

$3\displaystyle\frac{1}{6}$

$3\displaystyle\frac{3}{10}$

$2\displaystyle\frac{1}{18}$

Solution 2: Problem analysis and strategy decision

The strategy must be to evaluate both the fraction expressions individually and then divide. Objective is then, how fast we can evaluate the two fraction expressions.

The patterns in both the expressions we discover is the ease of combining the denominators of the positive term pairs and negative term pairs separately. This is the key pattern in this problem.

Solution 2: Problem solving execution using key pattern

The positive pair of terms in the first and second expressions evaluate respectively to, $\displaystyle\frac{7}{10}$ and $1$.

Problem 3.

Solution 3: Problem solving execution

Taking care of the decimal point in the denominator of the second term, we evaluate the result of the division as,

$\displaystyle\frac{100}{2}=50$.

Subtracting from the first term we get the final result as,

$3034-50=2984$.

Answer: Option b. 2984.

Key conceptys used: Decimal elimination technique -- Quick solution.

Problem 4.

Value of $\displaystyle\frac{\displaystyle\frac{5}{3}\times{\displaystyle\frac{7}{51}}\text{ of }\displaystyle\frac{17}{5}-\displaystyle\frac{1}{3}}{\displaystyle\frac{2}{9}\times{\displaystyle\frac{5}{7}}\text{ of }\displaystyle\frac{28}{5}-\displaystyle\frac{2}{3}}$ is,

$\displaystyle\frac{1}{2}$

$4$

$\displaystyle\frac{1}{4}$

$2$

Solution 4: Problem analysis and solving execution

Knowing that the $\text{of}$ operator is equivalent to multiplication we evaluate first the product terms in the numerator and denominator. Cancelling out the common factors the products evaluate for the numerator and denominator respectively to,

Problem 7.

The value of $8\displaystyle\frac{1}{2}-\left[3\displaystyle\frac{1}{4}\div{\left\{1\displaystyle\frac{1}{4}-\displaystyle\frac{1}{2}\left(1\displaystyle\frac{1}{2}-\displaystyle\frac{1}{3}-\displaystyle\frac{1}{6}\right)\right\}}\right]$ is

This is an evaluation problem of mixed fraction expression guided by BODMAS rule.

BODMAS rule

General evaluation sequence: from left to right (LTR, just like reading sequence of a sentence in majority of natural languages). In an expression comprising of equal precedence operations only, this rule comes into play.

O: Orders or powers: We also know this as exponentials or indices. Example, $2^2\times{3}$. This has lower precedence than brackets but higher precedence than Division, Multiplication, Addition, Subtraction. Occasionally Overline, a line over a few consecutive terms is used for separating out the terms into an expression, with an effect like a first bracket.

D M: Division and Multiplication: these have higher precedence than only Addition and Subtraction. Whichever of Division and Multiplication occurs first from left to right sequence will be evaluated first. Occasionally, $\text{ Of }$ operator is used for multiplication.

A S: Addition and Subtraction: these are of lowest precedence. Whichever of the two occurs first in LTR sequence will be evaluated first.

Conventional approach would be to convert the mixed fractions to the equivalent improper fractions (for example, conversion of $1\displaystyle\frac{1}{2}$ to $\displaystyle\frac{3}{2}$) and proceed with BODMAS rule guided evaluation.

In mixed fraction expression evaluation, we don't convert mixed fraction terms to the equivalent improper fraction form. Instead, we consider each as a two part number: integer plus fraction. Consequently the expression is converted into an integer expression part and a pure proper fraction expression part, speeding up the overall process significantly.

For example, we consider $3\displaystyle\frac{1}{2}-2\displaystyle\frac{1}{3}$ as,

Solution 7: Problem solving execution

First adding $-\displaystyle\frac{1}{3}$ to $-\displaystyle\frac{1}{6}$ we get $-\displaystyle\frac{1}{2}$. It cancels out the $\displaystyle\frac{1}{2}$ of $1\displaystyle\frac{1}{2}$, leaving $1$ as the outcome.

Next subtracting $\displaystyle\frac{1}{2}$ from $1\displaystyle\frac{1}{4}$ results in $\displaystyle\frac{3}{4}$ which divides $3\displaystyle\frac{1}{4}=\displaystyle\frac{13}{4}$ to result in $\displaystyle\frac{13}{3}$.

Finally, this value of $\displaystyle\frac{13}{3}$ is subtracted from $8\displaystyle\frac{1}{2}=\displaystyle\frac{17}{2}$ to leave $\displaystyle\frac{25}{6}=4\displaystyle\frac{1}{6}$ as the answer.

We have not shown the deduction steps as the need is to evaluate mentally and quickly which is easily met by the process adopted.

Mark that near the end, we have converted mixed fractions to the equivalent improper fraction form to easily carry out first division and then subtraction. This is what we call flexible method. We changed the process slightly according to the need.

Solution 8: Problem analysis

We feel the problem needs a bit of careful calculation and decided to evaluate the second term of the division operation first and then the numerator and denominator of the first term.

We also note the key pattern of LCM of fraction denominators 28 in the numerator and 14 in the denominator of the first term. This should help in simplification. We will multiply the numerator and denominator by 28 to eliminate fractions of this term at the last stage.

Solution 8: Evaluation of the continued fraction

The last fraction expression evaluates to, $5-\displaystyle\frac{1}{5}=\displaystyle\frac{24}{5}$. By immediate inversion it is transformed to $\displaystyle\frac{5}{24}$ and the expression to be evaluated turns to,

$2+\displaystyle\frac{5}{24}=\displaystyle\frac{53}{24}$, and after inversion again, $\displaystyle\frac{24}{53}$. The next expression to be evaluated turns to,

$2+\displaystyle\frac{24}{53}=\displaystyle\frac{130}{53}$, and after inversion again finally,

Problem 9.

The value of $\displaystyle\frac{1+\displaystyle\frac{1}{2}}{1-\displaystyle\frac{1}{2}}\div{\displaystyle\frac{4}{7}\left(\displaystyle\frac{2}{5}+\displaystyle\frac{3}{10}\right)}\text { of }\displaystyle\frac{\displaystyle\frac{1}{2}+\displaystyle\frac{1}{3}}{\displaystyle\frac{1}{2}-\displaystyle\frac{1}{3}}$ is,

$37\displaystyle\frac{1}{2}$

$18\displaystyle\frac{3}{8}$

$\displaystyle\frac{3}{2}$

$\displaystyle\frac{2}{3}$

Problem analysis and solving execution

There are three main fraction terms separated by first, a division and second, a multiplication operation (the "$\text {of}$" operation is equivalent to multiplication). We will evaluate these three terms sequentially.

Mark that because of equal precedence of division and multiplication in BODMAS rule, and as expressions are evaluated from left to right, the division operation appears first and inverts $\displaystyle\frac{2}{5}$ to $\displaystyle\frac{5}{2}$ and only then the multiplication by 5 comes into play.

This is a good example of how left to right sequence of evaluation determines effective precedence in case of two operations of equal precedence according to BODMAS rule apppearing one after the other in an expression.